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**HallsofIvy** Plato asked if you could prove f(0)= 0, then gave a proof himself. You responded to that with

No, Plato meant nothing of the sort- he certainly never said anything about the function "F(x)= x- 2" because that is NOT an odd function! Further he was not talking about two **different** functions which is what you are doing.

A function is odd if, for all x, f(-x)= -f(x). That is not true for F(x)= x- 2: Taking x= 3, F(3)= 3- 2= 1. Taking x= -3, F(-3)= -3- 2= -5. $\displaystyle F(-x)\ne-F(x)$.

What Plato was saying was- start from f(-x)= -f(x) and let x= 0.

To show that a function, f, is continuous at x= a, we would have to prove that $\displaystyle \lim_{x\to a} f(x)= f(a)$. Here, we have a= 0 and we know that f(0)= 0 so that reduces to $\displaystyle \lim_{x\to 0} f(x)= 0$. We are also told that f is "right continuous at x= 0" which means $\displaystyle \lim_{x\to 0^+} f(x)= 0$ (that is the limit as we approach 0 only from the positive numbers). Now you know that f(-x)= -f(x).